Given a positive integer n, check if it is perfect square or not using only addition/subtraction operations and in minimum time complexity.
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We can use the property of odd number for this purpose:
Addition of first n odd numbers is always perfect square 1 + 3 = 4, 1 + 3 + 5 = 9, 1 + 3 + 5 + 7 + 9 + 11 = 36 ...
Below is the implementation of above idea :
How does this work?
Below is explanation of above approach.
1 + 3 + 5 + ... (2n-1) = ∑(2*i - 1) where 1<=i<=n = 2*∑i - ∑1 where 1<=i<=n = 2n(n+1)/2 - n = n(n+1) - n = n2
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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